Question

3-6. A drum contains 0.16 mº of toluene. If the lid is left open (lid diameter is 0.92 m), deter-

Answer 1

This problem is providing us with the volume of toluene, 0.16 m³, and the local ventilation rate 28.34 m³/min,  that is inside a drum at 30 °C and 1 atm. Thus, the time to evaporate all of the toluene is required as well as the parts per million in the drum, and found to be 96.2 min and 3.77 ppm respectively.

Ventilation:

In mass transfer problems, ventilation involves mass transfer coefficients, cross-section area and saturation pressure, that we can calculate for this problem as shown below:

$A=\frac{\pi}{4}*0.92m^2=0.665 m^2\\\\k=0.83\frac{cm}{s} (\frac{18.02g/mol}{92.13g/mol} )^{\frac{1}{3} }=0.482\frac{cm}{s}=0.00482\frac{m}{s}\\ \\ P_{sat}=exp(16.0137-\frac{3096.52}{53.67+303} )mmHg*\frac{1atm}{760mmHg}=2.01 atm$

After that, we can calculate the rate at which the toluene is evaporated:

$Qm=\frac{M_{tol}*k*A*P_{sat}}{RT}\\ \\Qm=\frac{(92.13\frac{g}{mol})(0.00482\frac{m}{s})(0.665m^2)(2.01atm)*\frac{1000L}{1m^3} }{(0.08206\frac{atm*L}{mol*K} )(303K)}=23.77\frac{g}{s}$

Where all mol, m, atm, L and atm are cancelled out. Next, given the volume of toluene and its density at 30 °C, 0.8574 g/mL, one can calculate its mass and hence the time to evaporate it all:

$m_T=0.16m^3*\frac{1x10^6 mL}{1m^3} *0.8574\frac{g}{mL}=137,184g$

$t=\frac{137,184g}{23.77\frac{g}{s} } =5771.3 s*\frac{1min}{60s} \\\\t=96.2min$

Next, the concentration of toluene in parts per million can be calculated with the rate at which toluene is evaporated and the local ventilation rate via the following equation:

$Cppm=\frac{Qm*R*T}{K*Qv*P*M_T} \\\\Cppm=\frac{(23.77\frac{g}{s}*\frac{1min}{60 s} )(0.08206\frac{atm*L}{mol*K} )(303K)}{(28.34\frac{m^3}{min}*\frac{1000L}{1m^3} )(1atm)(92.13\frac{g}{mol} )} *10^6\\\\Cppm=3.77ppm$

Learn more about mass transfer: brainly.com/question/25309236